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Question

# An electron moving along positive X-axis with a 3×106m/s speed, enters the region of a uniform electric field. The electron stops after travelling a distance of 90mm in the field. The electric field strength is (charge on electron =1.6×10−19, mass of electron=9.1×10−31kg) :

A
2.84KVm1 along -ve X-axis
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B
0.284KVm1 along -ve X-axis
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C
0.284KVm1 along +ve X-axis
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D
28.4KVm1 along +ve X-axis
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Solution

## The correct option is D 0.284KVm−1 along +ve X-axise stops after travelling a distance dNow work done by electric field must be equalto change in kinetic rengy of e−∴Eed=12mv2⇒E=12 (me)v2d⇒E=12 (9.1×10−311.6×10−19) (9×1012)90×10−3⇒ E=0.284×103∴ E=0.284KVm(^x)→ as e− is deaccelerating, force on it must be in opposite direction of velocity ∴ E is in ^x direction

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